Sequence of measures and a non-negative definite matrix Let $\mu_n$, $n \in \mathbb{N}$, be measures on $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d)$ such that $\mu_n \{0\}=0$
$$
\int_{|x| \leq \varepsilon} |x|^2 \mu_n (dx) < \infty, \quad \mu_n( \{ x \in \mathbb{R}^d : |x| > \varepsilon\} ) < \infty \quad \forall \varepsilon>0, \tag{1}
$$
Let $A = (A_{ i, j})_{i,j=1,\dots,d} \in \mathbb{R}^{d \times d}$ be a non-negative definite matrix and assume that
$$
\lim_{\varepsilon \downarrow 0} \limsup_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon} \langle y, x \rangle^2 \mu_n (dx)  - \langle y, A y 
 \rangle \right| = 0 \quad \forall y \in \mathbb{R}^d. \tag{2}
$$

How can one show that $$\lim_{ \varepsilon \downarrow 0} \limsup_{ n \rightarrow \infty } \int_{|x| \leq \varepsilon} x_i x_j \mu_n ( d x ) = \lim_{ \varepsilon \downarrow 0} \liminf_{ n \rightarrow \infty } \int_{|x| \leq \varepsilon} x_i x_j \mu_n ( d x ) = A_{ij} \quad ? \tag{3}$$

If necessary, one can assume that $\mu_n = a_n \lambda_n$ where $a_n \rightarrow \infty$ as $n \rightarrow \infty$, and $\lambda_n$ are finite (probability) measures.

Some observations:
Since $A$ is non-negative definite, $\langle y A, y \rangle \geq 0$ in $(2)$.
$(2)$ also implies that
$$
\limsup_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon} \langle y, x \rangle^2 \mu_n (dx)  - \langle y, A y 
 \rangle \right| < \infty.
$$
Since
$$
\liminf_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon} \langle y, x \rangle^2 \mu_n (dx)  - \langle y, A y 
 \rangle \right| \leq \limsup_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon} \langle y, x \rangle^2 \mu_n (dx)  - \langle y, A y \rangle \right|
$$
it also follows from $(2)$ that
$$
\lim_{\varepsilon \downarrow 0} \liminf_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon} \langle y, x \rangle^2 \mu_n (dx)  - \langle y, A y 
 \rangle \right| = \lim_{\varepsilon \downarrow 0} \limsup_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon} \langle y, x \rangle^2 \mu_n (dx)  - \langle y, A y \rangle \right| = 0.
$$
To show $(3)$, let us take $y = (y_1, \ldots, y_d) \in \mathbb{R}^d$ such that $y_i=y_j=1$, $i \neq j$, and the remaining components are $0$. Then
$$
\langle y, A y \rangle = \sum_{k,m = 1}^d A_{ k m } y_m y_k = A_{i j} y_i y_j + A_{j i} y_j y_i + A_{ii} y_i^2 + A_{jj} y_j^2= 2 A_{i j} + A_{ii} +A_{jj} \geq 0.
$$
From $(2)$ we then get
$$
\lim_{\varepsilon \downarrow 0} \limsup_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon}  (x_i + x_j)^2 \mu_n (dx)  - 2A_{ij} - A_{ii} - A_{jj} \right|= 0
$$
$\Longrightarrow$
$$
\lim_{\varepsilon \downarrow 0} \limsup_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon}  (x_i^2 + x_j^2 ) \mu_n (dx) -A_{ii} - A_{jj} + 2 \int_{|x| \leq \varepsilon} x_i x_j \mu_n (dx) - 2 A_{ij} \right|= 0.
$$
How can one proceed from here? 
 A: Let $i \in \{1, \ldots, d \}$, $y = (y_1, \ldots, y_d)$ with $y_i = 1$ and let the remaining components be $0$. Then
$$
\langle y, A y \rangle = \sum_{k,m = 1}^d A_{ k m } y_m y_k = A_{i i} y_i y_i = A_{ii},
$$
and from $(2)$ we get
$$
\lim_{\varepsilon \downarrow 0} \limsup_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon} x_i^2 \mu_n (dx)  - A_{ii} \right| = 0. \tag{4}
$$
Now let $i, j \in \{1, \ldots, d\}$, $i \neq j$, $y= (y_1, \ldots, y_n)$ with $y_i=y_j=1$ and let the remaining components be $0$. Then
$$
\langle y, A y \rangle = \sum_{k,m = 1}^d A_{ k m } y_m y_k = A_{i j} y_i y_j + A_{j i} y_j y_i + A_{ii} y_i^2 + A_{jj} y_j^2= 2 A_{i j} + A_{ii} +A_{jj},
$$
and from $(2)$ we get
$$
\lim_{\varepsilon \downarrow 0} \limsup_{n \rightarrow \infty} \left| \int_{|x| \leq \varepsilon}  (x_i^2 + x_j^2 ) \mu_n (dx) -A_{ii} - A_{jj} + 2 \int_{|x| \leq \varepsilon} x_i x_j \mu_n (dx) - 2 A_{ij} \right|= 0.
$$
Then
\begin{align*}
\left| 2 \int_{|x| \leq \varepsilon} x_i x_j \mu_n (dx) - 2 A_{ij} \right| &\leq \left| \int_{|x| \leq \varepsilon}  (x_i^2 + x_j^2 ) \mu_n (dx) -A_{ii} - A_{jj} + 2 \int_{|x| \leq \varepsilon} x_i x_j \mu_n (dx) - 2 A_{ij} \right| \\
&\quad + \left| \int_{|x| \leq \varepsilon} x_i^2 \mu_n (dx)  - A_{ii} \right| + \left| \int_{|x| \leq \varepsilon} x_j^2 \mu_n (dx)  - A_{jj} \right|,
\end{align*}
hence
$$
\lim_{\varepsilon \downarrow 0} \limsup_{n \rightarrow \infty} \left| 2 \int_{|x| \leq \varepsilon} x_i x_j \mu_n (dx) - 2 A_{ij} \right|= 0. \tag{5}
$$
$(4)$ and $(5)$ together imply $(3)$.
